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Arte-miy333 [17]
3 years ago
7

Assume that you measured exactly 2.0 mL of the sodium chloride solution into a small test tube. How many moles of sodium chlorid

e are in the tube?
Chemistry
2 answers:
Lera25 [3.4K]3 years ago
4 0

Answer:

0.034 mol (4dp)

Explanation:

Use the formula n = m/M where n= number of moles, m=mass in grams and M=molar mass in g/mol.

Therefore to find how many moles, divide the mass in grams (2) by the molar mass of sodium chloride which is 22.99+35.45

ruslelena [56]3 years ago
3 0
0.034 moles (rounded answer)

The molar mass of sodium chloride is 58.44 g/mol. Therefore, if you’re given 2.0 mL of it, that would be 0.034 moles.
2.0/58.44 = 0.03422313
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How many molecules are in 100 g of C6H120,?*​
Grace [21]

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

6 0
3 years ago
Below is a picture of two blocks made of different materials. Block A Block B grams grams If Block A is located on the Moon and
AURORKA [14]

Answer:

A

Explanation:

this is because density is the mass per unit volume of an object

4 0
3 years ago
When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced.
sp2606 [1]

Answer:

Mass of CaCl₂ =  20 g

CaCO is presewnt in excess.

Mass of of CaCO₃ remain unreacted =  7.007 g

Explanation:

Given data:

Mass of calcium carbonate = 25 g

Mass of hydrochloric acid = 13.0 g

Mass of calcium chloride produced = ?

Chemical equation:

CaCO₃ + 2HCl  →  CaCl₂  + H₂O + CO₂

Number of moles of CaCO₃:

Number of moles of CaCO₃ = Mass /molar mass

Number of moles of CaCO₃= 25.0 g / 100.1 g/mol

Number of moles of CaCO₃ = 0.25 mol

Number of moles of HCl:

Number of moles of  HCl = Mass /molar mass

Number of moles of HCl = 13.0 g / 36.5 g/mol

Number of moles of HCl = 0.36 mol

Now we will compare the moles of CaCl₂ with HCl and CaCO₃ .

                  CaCO₃         :               CaCl₂

                    1                 :               1

                 0.25              :            0.25

                HCl                :                CaCl₂

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

The number of moles of CaCl₂ produced by HCl are less it will be limiting reactant.

Mass of CaCl₂ = moles × molar mass

Mass of CaCl₂ =0.18 mol × 110.98 g/mol

Mass of CaCl₂ =  20 g

The calcium carbonate is present in excess.

                HCl                :                CaCO₃

                 2                   :                    1

                 0.36            :                  1/2 × 0.36 = 0.18 mol

So, 0.18 moles react with 0.36 moles of HCl.

The moles of CaCO₃ remain unreacted = 0.25 -0.18

The moles of CaCO₃ remain unreacted = 0.07 mol

Mass of of CaCO₃ remain unreacted = Moles × molar mass

Mass of of CaCO₃ remain unreacted = 0.07 mol × 100.1 g/mol

Mass of of CaCO₃ remain unreacted =  7.007 g

7 0
3 years ago
A student wants to make a 0.600 M aqueous solution of barium sulfate, BaSO4, and has a bottle containing 12.00 g of barium sulfa
love history [14]

<u>Answer: </u>The volume of the solution is 85.7 mL

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

\text{Molarity of solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (mL)}} .....(1)

We are given:

Molarity of solution = 0.600 M

Given mass of BaSO_4 = 12.00 g

We know, molar mass of BaSO_4=[(1\times 137.33)+(1\times 32.07)+(4\times 16)]=233.4g/mol

Putting values in equation 1, we get:

0.600=\frac{12.00\times 1000}{233.4\times \text{Volume of solution}}\\\\\text{Volume of solution}=\frac{12.00\times 1000}{233.4\times 0.600}=85.68mL=85.7mL

The rule of significant number that is applied for the problems having multiplication and division:

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

Here, the least number of significant figures is 3 that is determined by the number, 0.600. Thus, the answer must have these many significant figures only.

Hence, the volume of the solution is 85.7 mL

5 0
3 years ago
Please helpppppppppppppppp!!!!!!!!!!!!!!!
Agata [3.3K]

Answer:

c MgO

Explanation:

product is located after the reaction arrow

8 0
3 years ago
Read 2 more answers
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